how can i solve integral equation ,symbolically methode by using a matlab

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aissa ghaddab 2021-7-16

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评论： Walter Roberson 2021-7-17

y(x)+ integral(k(x,t)*y(t) , t , a, b) =f(x)

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Answer 1

Sulaymon Eshkabilov 2021-7-16

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You can start by initiating the variables via syms, e.g.:

syms t x
y = expression1;
f = expression2;
Sol = solve(y+integral(@(t)expression3)-expression2==0)

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Answer 2

Walter Roberson 2021-7-16

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syms a b f(x) k(x,t) y(x)

eqn = y(x) + int(k(x,t)*y(t), t, a, b) == f(x)

eqn =

But what are you trying to solve for? For example do you have known f(x) and y(x) and a and b, and the task is to find k(x,t) ? Or do you have known functions and known a and the task is to find b?

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aissa ghaddab 2021-7-16

actually, I wanna find y(x),

for example:

we give f(x)=exp(-x^2) , a=0, b=1, k(x,t)=exp(x*t)

Walter Roberson 2021-7-17

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That appears to be an inhomogeneous Fredholm equation of the second kind; it looks like it would typically be examined using Liouville–Neumann series https://en.wikipedia.org/wiki/Liouville%E2%80%93Neumann_series . For the usual approach to hold,

syms x t a b real

Kcond = int(abs(exp(x*t)),t,a,b)

Kcond =

which is (exp(b*x)-exp(a*x))/x .

This can be true for particular ranges of x, but clearly cannot be true for arbitrary real a, b, and x ranges. Picking a and b arbitrarily,

Kcond13 = subs(Kcond, [a,b], [1, 3])

Kcond13 =

vpasolve(Kcond13 == 1)

ans =

limit(Kcond13, x, -inf)

ans =

0

so it could be true from -inf to about -0.357 but not for greater x.

What this tells us is that the Liouville Neumann approach is not suitable for this equation -- that there is no general contraction mapping.

No matter which non-zero a and b you choose, there is some x for which (exp(b*x)-exp(a*x) is arbitrarily large, and that increases non-linearly, so at some point that is going to be greater than x. So it looks to me as if that particular set of equations does not have a contraction mapping, and so probably there is no solution.

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